I am given the following system of linear equations:
$$
\left\{
\begin{array}{c}
2x+2y+mxy=5 \\
(m-1)(x+y)+xy=1 \\
3x+3y-xy=m+1
\end{array}
\right.
$$
where $m \in \mathbb{R}.$
I am asked to find the number of values of $m$ (so not the values themselves, just the number of values) for which the system has solutions $(x,y) \in \mathbb{R} \times \mathbb{R}$. I tried a bunch of things.
First, I tried performing gaussian elimination on the system using the system like so:
$$
\begin{bmatrix}
2 & 2 & m \\
m-1 & m-1 & 1 \\
3 & 3 & -1 \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
xy \\
\end{bmatrix}
=
\begin{bmatrix}
5 \\
1 \\
m+1 \\
\end{bmatrix}
$$
But I got stuck. I saw that $2^{nd} column = 3^{rd} column$ so the determinant of that matrix is $0$. So the ranks of the matrices $A$ and $\overline{A}$ (augmented $A$) must be the same, smaller than $3$ (since $det(A)=0$). I thought I could use this fact but couldn't come up with anything.
Then I tried factoring a $2$ in the first equation and a $3$ in the last equation and came up with this other system:
$$
\begin{bmatrix}
2 & m\\
m-1 & 1\\
3 & -1\\
\end{bmatrix}
\begin{bmatrix}
x+y \\
xy \\
\end{bmatrix}
=
\begin{bmatrix}
5 \\
1 \\
m+1 \\
\end{bmatrix}
$$
I tried similar things here but I still couldn't find a solution.
Any help would be appreciated.
PS: I know from the source of this problem that the answer is $1$ (so the number of real values of $m$ for which the system has solutions $(x,y) \in \mathbb{R} \times \mathbb{R}$ is $1$).
Best Answer
Because of $$ \underbrace{ \begin{bmatrix} 2 & m & 5\\ m\color{red}{-1} & 1 & 1\\ 3 & -1 &m+1 \end{bmatrix}}_{=:A} \begin{bmatrix} x+y \\ xy \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0 \end{bmatrix} \qquad(*) $$ the possible values of $m$ correspond to the roots of the determinant of the matrix denoted $A$ above. This determinant is zero iff $$ (m^2 - 2m + 3)(m + 2) =0\ . $$ Since we are working over reals, only $m=-2$ has to be studied.
Later edit: I am trying to answer the question in the comments part, doing things as detailed as i can. Let $m$ be a value that accepts a solution $(x,y)$ of the given system of algebraic equations of degree two. We have then an associated homogeneous linear system $(*)$ above with a non-trivial solution. (That $-1$ makes it non-trivial.) (Alternatively, in the school they claim that an overdetermined system has a solution if the rank of the "extended matrix", in our case $A$, coincides with the rank of the matrix of the system. The latter rank is at most two, so we have an other argument for getting $\det A=0$.) This implies that the matrix of the system, denoted by $A$, is singular. This implies that $\det A=0$ (for this specific $m$). We compute this determinant (linear operations on the rows or Sarrus) and factorize. The only real possible value for $m$ is $m=-2$.
(This was one implication, we used the given condition to restrict the possible values of $m$, now we have to consider "them" in particular, this only possible value $m=-2$, for the other implication.)
So we have to look then specifically, to see if $m=-2$ really leads to a real solution. The system in the unknown $s$ (sum, $s=x+y$), and $p$ (product, $p=xy$), becomes for $m=-2$: $$ \begin{bmatrix} 2 & -2 & 5\\ -3 & 1 & 1\\ 3 & -1 &-1 \end{bmatrix} \begin{bmatrix} s \\ p \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0 \end{bmatrix}\ , $$ we get $s=-7/4$, $p=-17/4<0$, and since $p<0$ we have (two) real solutions (for this one value of $m$).